Here is the best I was able to do on puzzle 2 (along with my reasoning):
The prime factors of 2022 are 2, 3, and 337. Any method of selecting 1 person from 2022 must cut the space down by a factor of 2, and by a factor of 3, and by a factor of 337 (it does not need to be in that order and you can filter down by more than one of those factors in single roll, but you must filter down by each of those in a way where the probability is uniform before starting).
The lowest it could be is 2 rolls. If someone could win on the first roll, that person’s probability of winning could be no less than 1/(Number of sides of the first roll die). Since the die with the most sides has 2017, that person’s probability to win would be more than 1/2022, so the probability of winning could not be even for everyone.
To get it in 2 rolls:
Before the start of the dice rolling, divide the group of 2022 using 3 different groupings:
Grouping A: Divide the 2022 people into 674 sub-groups of 3 people each (Group A1, Group A2, … Group A674)
Grouping B: Divide the 2022 people into 1011 sub-groups of 2 people each (Group B1, Group B2, … Group B1011)
Grouping C: Divide the 2022 people into 6 sub-groups of 337 people each—but differentiated by 0-indexed numbers that correspond to modulo amounts (Group C0, Group C1, Group C2, Group C3, Group C4, and Group C5)
Each person will be a member of exactly 1 A group, exactly 1 B group, and exactly 1 C group.
For the first roll, roll the die with 1697 sides.
If the number is between 1 and 674 (inclusive):
Select the A group whose number corresponds to the number of the die.
Roll the 3-sided die to select a winner from among that group.
If the number is between 675 and 1685 (inclusive):
Calculate: ((Number on the die) − 674) to get a number between 1 and 1011 (inclusive)
Select the B group whose number corresponds to the ((Number on the die) − 674) number.
Roll the 2-sided die to select a winner from among that group.
If the number is between 1686 and 1697 (inclusive):
Calculate: (((Number on the die) − 1685) modulo 6) to get a number between 0 and 5 (inclusive)
Select the C group whose number corresponds to the (((Number on the die) − 1685) mod 6) number.
Roll the 337-sided die to select a winner from among that group.
So on this calculation, the expected number of rolls is exactly 2, since for each possible outcome on the first one, there is a second die to throw that will select the winner.
Here is the best I was able to do on puzzle 2 (along with my reasoning):
The prime factors of 2022 are 2, 3, and 337. Any method of selecting 1 person from 2022 must cut the space down by a factor of 2, and by a factor of 3, and by a factor of 337 (it does not need to be in that order and you can filter down by more than one of those factors in single roll, but you must filter down by each of those in a way where the probability is uniform before starting).
The lowest it could be is 2 rolls. If someone could win on the first roll, that person’s probability of winning could be no less than 1/(Number of sides of the first roll die). Since the die with the most sides has 2017, that person’s probability to win would be more than 1/2022, so the probability of winning could not be even for everyone.
To get it in 2 rolls:
Before the start of the dice rolling, divide the group of 2022 using 3 different groupings:
Grouping A: Divide the 2022 people into 674 sub-groups of 3 people each (Group A1, Group A2, … Group A674)
Grouping B: Divide the 2022 people into 1011 sub-groups of 2 people each (Group B1, Group B2, … Group B1011)
Grouping C: Divide the 2022 people into 6 sub-groups of 337 people each—but differentiated by 0-indexed numbers that correspond to modulo amounts (Group C0, Group C1, Group C2, Group C3, Group C4, and Group C5)
Each person will be a member of exactly 1 A group, exactly 1 B group, and exactly 1 C group.
For the first roll, roll the die with 1697 sides.
If the number is between 1 and 674 (inclusive):
Select the A group whose number corresponds to the number of the die.
Roll the 3-sided die to select a winner from among that group.
If the number is between 675 and 1685 (inclusive):
Calculate: ((Number on the die) − 674) to get a number between 1 and 1011 (inclusive)
Select the B group whose number corresponds to the ((Number on the die) − 674) number.
Roll the 2-sided die to select a winner from among that group.
If the number is between 1686 and 1697 (inclusive):
Calculate: (((Number on the die) − 1685) modulo 6) to get a number between 0 and 5 (inclusive)
Select the C group whose number corresponds to the (((Number on the die) − 1685) mod 6) number.
Roll the 337-sided die to select a winner from among that group.
So on this calculation, the expected number of rolls is exactly 2, since for each possible outcome on the first one, there is a second die to throw that will select the winner.